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A teacher wanted to distribute 900 chocolates among the students of a class. Each boy received 12 chocolates and each girl received 6 chocolates. If each girl had been given 10 chocolates, then each boy would have received 5 chocolates. Find the number of students of the class. (1) 80 (2) 90 (3) 100 (4) 110

Short Answer

Expert verified
Answer: There are 110 students in the class.

Step by step solution

01

Define the variables

Let x be the number of boys and y be the number of girls in the class.
02

Understand the problem conditions and set up equations

According to the problem statement, each boy received 12 chocolates and each girl received 6 chocolates; This resulted in the distribution of 900 chocolates. So, we have the equation 12x + 6y = 900. Now, we are given another condition in which if each girl had been given 10 chocolates, then each boy would have received 5 chocolates. In this case, we have another equation: 5x + 10y = 900.
03

Solve the system of equations

We have two equations: 1. 12x + 6y = 900 2. 5x + 10y = 900 First, divide both sides of equation 1 by 6 to simplify it: 2x + y = 150 (Equation 1') Now we will multiply equation 1' by 10 in order to make the coefficients of y the same in both equations: 20x + 10y = 1500 Next, subtract equation 2 from this equation: 20x + 10y - 5x - 10y = 1500 - 900 15x = 600 x = 40 Now that we have the number of boys (x), we can substitute it into equation 1' to find the number of girls (y): 2(40) + y = 150 80 + y = 150 y = 70
04

Find the number of students

Now that we have found the number of boys (x = 40) and the number of girls (y = 70), we can find the total number of students in the class by adding the two numbers together: Total number of students = x + y = 40 + 70 = 110
05

Choose the correct answer

The correct answer is (4) 110, as there are a total of 110 students in the class.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chocolate Distribution Problem
The chocolate distribution problem is an interesting exercise that involves distributing a fixed number of chocolates among boys and girls in a classroom. It is an example of a real-world scenario that can be solved using a system of equations. The challenge here is to find out how many boys and girls are in the class based on the total number of chocolates given and under different distributions for boys and girls.

By setting up and solving these equations, students will learn how algebraic methods can help quantify unknowns in practical problems. The conditions initially state that each boy receives 12 chocolates while each girl receives 6 chocolates, resulting in a total distribution of 900 chocolates. A second scenario shifts the distribution where each girl gets 10 chocolates, causing each boy to receive 5 chocolates, but still maintaining the total at 900 chocolates.

This problem is ideal for practicing the setup and solution of systems of equations, understanding constraints, and observing how changes in distribution affect variables.
Algebraic Problem-Solving
Algebraic problem-solving is the process of representing problems with mathematical symbols and expressions, and then using algebraic techniques to find their solutions. In the chocolate distribution problem, algebra is used to translate the problem's conditions into a mathematical framework that can be analyzed and solved.

The main tool here is a system of linear equations, where each equation corresponds to a distinct condition or scenario related to the problem. These equations must be expressed in terms of unknown variables, such as the number of boys and girls in the class.

To solve these equations, methods like substitution or elimination are employed. These techniques involve combining, rearranging, and simplifying equations to isolate the variables, thus finding their values. By practicing with problems like the chocolate distribution, you can become adept at turning word problems into solvable mathematical expressions.
Variables in Mathematics
In mathematics, variables are symbols used to represent unknown values. They are fundamental in formulating equations and solving various mathematical problems. In the chocolate distribution scenario, we use variables to denote unknowns: \(x\) for the number of boys and \(y\) for the number of girls.

Using variables allows for a flexible approach to problem-solving, as they can take on different values depending on how the conditions or parameters of a problem are set. When solving algebraic equations, our goal is to determine the values of these variables that satisfy all given conditions.

Variables help keep track of quantities when conditions or coefficients change, as seen in the alternative chocolate distribution scenario. Becoming comfortable with using variables is key in transitioning from arithmetic to more advanced mathematical topics like algebra.
Linear Equations
Linear equations are fundamental building blocks in algebra. They are called 'linear' because they graph as straight lines in a two-dimensional space. A linear equation contains variables, constants, and operations of addition, subtraction, multiplication, or division, and is set to equal a constant.

In the context of the chocolate distribution problem, two linear equations were developed to model the distribution of 900 chocolates under different scenarios. The first equation 12\(x\) + 6\(y\) = 900 represents the initial distribution, while the second, 5\(x\) + 10\(y\) = 900, represents the alternate scenario.

Solving a system of linear equations requires finding a common solution that satisfies all equations simultaneously. Typically, equations can be simplified or transformed to highlight relationships between variables. For instance, manipulating the equations through methods like substitution or elimination helped reveal that there are 40 boys and 70 girls, resulting in a total of 110 students.

Mastering linear equations provides a foundation for tackling more complex algebraic concepts and real-world problem-solving scenarios.

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Most popular questions from this chapter

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